Question

What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? ( Mass of earth is 6 × 10^24 kg and radius of earth is 6.4 × 10^6 m)

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Answer 1

Given:-

• $\sf \: G = 6.7 × 10^{-11} \: Nm²/kg²$
• $\sf \: m_1= 6×10^{24} \: kg$
• $\sf \: m_2 = 1 \: kg$
• $\sf \: r = 6.4× 10^{6} \: m$

★ Using formula, we get

$\sf \: F = G × \frac{m_1 \times m_2}{ {r}^{2} }$

$\sf \longrightarrow F = \frac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{(6.4 \times {10}^{6})^{2} }$

$\sf \longrightarrow F = \frac{6.7 \times {10}^{ - 11} \times 6 \times {10}^{24} \times 1 }{6.4 \times {10}^{6} \times 6.4 \times {10}^{6} }$

$\sf \longrightarrow F = \frac{6.7 \times 6 \times 1 \times {10}^{ - 11} \times {10}^{24} \times {10}^{ - 12} \times 10 }{6.4 \times 6.4}$

$\sf \longrightarrow F = \frac{67 \times 6 \times 10 \times 10}{64 \times 64}$

$\sf \longrightarrow F = \frac{4020 \times {10}^{1} }{4090}$

$\sf \longrightarrow F = 0.98145 \times {10}^{1}$

$\sf \longrightarrow F = 9.8 \: m/ {s}^{2}$

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Thus, the magnitude of gravitational force between the earth and a 1 kg object on its surface is 9.8 Newtons.

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Answer 2

Given:-

• Mass of earth m1 =6 × 1024 kg
• Mass of the object m2 = 1 kg
• The radius of the earth =6.4 × 106 m
• Universal gravitational constant G = 6.67 x 10-11 Nm2kg-2

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Find out:-

• The magnitude of the gravitational force between the earth and a 1 kg object on its surface

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Solution:-

By applying universal law of gravitation: – F = (G m1 m2/r2)

= (6.67 x 10-11 x 6 × 1024 x 1)/ (6.4 × 106)2

F = 9.8 N

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Answer:-

The magnitude of the gravitational force between the earth and a 1 kg object on its surface is F= 9.8N

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